CONGRUENCES MODULO 4 OF CALIBERS OF REAL QUADRATIC FIELDS
MASANOBUKANEKOANDKEITAMORI
Dedicated to professor Paulo Ribenboim on the occasion of his 80th birthday.
RÉSUMÉ. Nous présentons des conjectures sur les congruences modulo4des ca- libres (au sens restreint) d’ordres quadratiques réels pour certains discriminants, et nous donnons des preuves dans quelques cas.
ABSTRACT. We present conjectures on congruences modulo 4 of the calibers (in the narrow sense) of real quadratic orders of particular discriminants, and prove some partial results on them.
1. Definitions and results
A real quadratic number w isreduced if it satisfies w > 1 and−1 < w0 < 0, wherew0 is the algebraic conjugate ofw over the rationalsQ. It is well known that wis reduced if and only if its usual continued fraction expansion is purely periodic.
Consider the set of all reduced quadratic numbers of a given discriminantD:
Q(D) :=
w|disc(w) =D, w >1, −1< w0<0 .
Here, a real quadratic irrationalitywis of discriminantD, denoteddisc(w) =D, ifw satisfies
aw2+bw+c= 0, a, b, c∈Z, a >0, GCD(a, b, c) = 1, b2−4ac=D . The setQ(D)is finite and its cardinality is denoted byκ(D). WhenDis a fundamental discriminant,i.e., the discriminant of the real quadratic fieldQ(√
D), the numberκ(D) is referred to as thecaliberofQ(√
D)(see [1]).
Any quadratic number of discriminantDis equivalent under the action ofGL2(Z) (via the linear fractional transformation) to an element inQ(D). We writew1 ∼ w2 if the two numbersw1 andw2 areGL2(Z)-equivalent. It is known that w1 ∼ w2 if and only if their periods of continued fraction expansions are cyclically equivalent. Let R(D)be the set ofGL2(Z)-equivalence classes ofQ(D)andh(D)be its cardinality:
R(D) =Q(D)/∼, h(D) =]R(D).
The numberh(D)is nothing but the wide class number of discriminantD.
Reçu le 22 février 2009 et, sous forme définitive, le 4 avril 2009.
We also consider the corresponding notions forSL2(Z)-equivalence. We call a real quadratic numberwm-reducedifw > 1and0 < w0 < 1. A number ism-reduced if and only if its “minus” continued fraction expansion is purely periodic (see [5, §13]).
LetQ+(D)be the set of allm-reduced numbers of a given discriminantD:
Q+(D) :=
w|disc(w) =D, w >1, 0< w0<1 .
This is also a finite set and its cardinality will be denoted byκ+(D). We refer to this number as them-caliberof (not necessarily fundamental) discriminantD. Consider the equivalence under the action ofSL2(Z). Two numbersw1 andw2 are strictly equiv- alent, writtenw1 ≈ w2, if the two are related with each other by a transformation in SL2(Z). Any quadratic number of discriminantDis strictly equivalent to an element inQ+(D), and two elements inQ+(D)are strictly equivalent if and only if the peri- ods of their minus continued fraction expansions are cyclically equivalent. LetR+(D) be the set ofSL2(Z)-equivalence classes ofQ+(D)andh+(D)be its cardinality (the
“narrow” class number):
R+(D) =Q+(D)/≈, h+(D) =]R+(D).
Throughout the paper, we denote byp a prime number congruent to 1 modulo 4, and byqa prime number larger thanpwhich is also congruent to 1 modulo 4. Integers xp, yp, xq, yqare uniquely defined by the decomposition
p=x2p+y2p, (0< xp < yp), q =x2q+y2q, (0< xq< yq). We denote byεDthe fundamental unit of discriminantDand byN(εD)its norm.
We conducted numerical experiments on the 2-orders of them-calibersκ+(8p)and κ+(pq), which are easily seen to be even integers, and observed the following1. Conjectures.(1)We have
κ+(8p)≡1−(−1)xp (mod 4).
(2)Assumep < qand supposexp 6≡xq (mod 2). Then we have κ+(pq)≡1−(−1)xp
q p
(mod 4).
For the narrow class numbers h+(8p) andh+(pq) (which are always even), the following two properties are known (cf.[2]):
(i) h+(8p)is divisible by 4 if and only ifp≡1 (mod 8), and (ii) h+(pq)is divisible by 4 if and only if
q p
= 1.
The above conjectures say that the divisibility by 4 ofκ+(8p)andκ+(pq)depends on theparityofxp orxq, not only on
q p
in the case (2), which may be of considerable interest.
1Special case of the conjecture and Theorem 1.1 were discussed in Umeno [4].
In this paper, we prove (1) in the case whereN(ε8p) =−1, and some partial result for (2) also in the case whereN(εpq) =−1.
Theorem 1.1. Letp be a prime number such thatp ≡ 1 (mod 4), and suppose N(ε8p) =−1. Then the class modulo 4 of them-caliberκ+(8p)is given by
κ+(8p)≡1−(−1)xp (mod 4).
As far as Conjecture (2) is concerned, we are not able to prove it even under the conditionN(εpq) = −1, but we obtain the following partial result. Notice that, corre- sponding to the two decompositions ofpqinto sums of two squares
pq= (xpxq+ypyq)2+ (xpyq−ypxq)2= (xpyq+ypxq)2+ (xpxq−ypyq)2, there exist exactly two quadratic numbersα, β ∈ Q(pq)whose periods of usual con- tinued fraction expansions are palindromic. We denote byl(α)andl(β)their period lengths.
Theorem 1.2. Besides the assumptionsp < q andxp 6≡ xq (mod 2), we further supposeN(εpq) =−1. Then we have
κ+(pq)≡1 + sgn(xpyq−ypxq)(−1)xp+l(α)−l(β)2 q
p
(mod 4).
Thus our conjecture in the caseN(εpq) =−1is that
xpyq−ypxq <0 if and only if l(α)≡l(β) (mod 4).
2. Preliminaries
As it is recalled in the previous section, the elementsα ∈ Q(D)andβ ∈ Q+(D) have purely periodic usual and minus continued fraction expansions
α= [a0, . . . , an−1] =a0+ 1
a1+ 1
. ..+ 1
an−1+ 1 a0+ 1
a1+ 1 . ..
and
β = [[b0, . . . , bm−1]] =b0− 1
b1− 1
. ..− 1
bm−1− 1 b0− 1
b1− 1 . ..
respectively. We definel(α) :=nandl+(β) :=mto be their minimum period lengths;
moreover, S(α) :=
n−1
X
i=0
ai and S+(β) :=]
j| 0≤j≤m−1, bj ≥3
are respectively called the sum of partial quotients and the number of partial quotients greater than or equal to3in the period.
In order to establish our results, we use the following result which may be standard, but for which, for the sake of convenience, we reproduce the proof given in Suzuki [3].
Proposition 2.1. We have κ+(D) = X
[α]∈R(D)
S(α) and κ(D) = X
[β]∈R+(D)
S+(β).
Proof. We prove the first assertion, the second being similarly proved (we shall not use the second formula). By definition, we have
κ+(D) = X
[β]∈R+(D)
l+(β).
We may assume that all representativesβare bigger than 2, because at least one of the partial quotientsbj in the expansion β = [[b0, . . . , bm−1]]is bigger than 2 and hence by a cyclic permutation we obtain a number greater than 2, equivalent toβ. As is easily seen, the map
(2.1) T : Q(D)3α 7→α+ 1∈ Q+(D)
gives a bijection between the sets Q(D) and {β ∈ Q+(D) | β > 2}.If the usual continued fraction expansion of α ∈ Q(D) is α = [a0, . . . , an−1], then the minus continued fraction expansion ofT(α) =α+ 1is given by (seee.g.[5])
[[a0+ 2,2, . . . ,2
| {z }
a1−1
, a2+ 2,2, . . . ,2
| {z }
a3−1
, . . . , an−2+ 2,2, . . . ,2
| {z }
an−1−1
]] ifnis even,
[[a0+ 2,2, . . . ,2
| {z }
a1−1
, . . . , an−1+ 2,2, . . . ,2
| {z }
a0−1
, . . . , an−2+ 2,2, . . . ,2
| {z }
an−1−1
]] ifnis odd.
Ifnis even, then the images underT of[ai, . . . , an−1, a0, . . . , ai−1]which are equiv- alent toαsplit into two classes (under the equivalence≈) according to the parity ofi.
By the formula above we have
l+ T([a0, . . . , an−1])
=a1+a3+· · ·+an−1, l+ T([a1, . . . , an−1, a0])
=a2+a4+· · ·+an−2+a0, and hence
l+ T([a0, . . . , an−1])
+l+ T([a1, . . . , an−1, a0])
=S(α). Ifnis odd, then allT([ai, . . . , an−1, a0, . . . , ai−1])are equivalent and
l+(T(α)) =S(α).
It is clear that the equivalenceT(α1) ≈ T(α2) impliesα1 ∼ α2. From this and the surjectivity of (2.1), we conclude the assertionκ+(D) =P
[α]∈R(D)S(α).
The following lemma is repeatedly used in our proof.
Lemma 2.2. Let
E = 0 1
1 0
and O= 1 1
1 0
be elements inGL2(F2)of order 2 and 3 respectively. Consider the product M =M1· · ·Mn
of lengthnofk O’s and(n−k)E’s (i.e.,Mi =EorO and the number ofO’s isk).
Then we have the equivalence n≡k (mod 2)⇐⇒M ∈
I =
1 0 0 1
, O = 1 1
1 0
, O2 = 0 1
1 1
.
Proof. Using relationsE2 =I, O3 =I, OEO =E(OE =EO2), any product ofE’s andO’s reduces to one of the six elements
I, O, O2, E, EO = 1 0
1 1
, OE = 1 1
0 1
. In this reduction process, the parity of
(total length of product)−(number ofO’s)
does not change, and out of the above six elements this parity is even exactly when it
isI, O, O2.
Lemma 2.3. Letα∈ Q(D).
(i) IfDis odd, then we havel(α)≡S(α) (mod 2).
(ii) IfDis even andN(εD) =−1, thenS(α)is even.
Proof. Let the quadratic equation satisfied byαbe
aα2+bα+c= 0 with GCD(a, b, c) = 1 and b2−4ac=D . For the continued fraction expansionα= [a0, a1, . . . , an−1], we set
p q r s
:=
a0 1 1 0
a1 1 1 0
· · ·
an−1 1
1 0
. Then we haveα= pα+q
rα+sand thus
rα2+ (s−p)α−q = 0.
Putd:= GCD(r, s−p, q) (>0). Then we haver =da,s−p=db, andq =−dc.
(i) AssumeDis odd. Thenbis odd. Ifdis odd, thens−pis odd and sopsis even, henceqris odd becauseps−qr= (−1)n. Thus
p q r s
≡ 1 1
1 0
=O or
0 1 1 1
=O2.
Ifdis even, thenr≡q ≡0, p≡s (mod 2), whereupon p q
r s
≡ 1 0
0 1
=I (mod 2).
By Lemma 2.2, we concluden≡S(α) (mod 2), because the parities ofS(α)and of the number of oddaiare the same.
(ii) AssumeDis even. Thenbshould be even and sop≡s (mod 2). Ifp≡s≡1 (mod 2)andq ≡ r ≡ 0 (mod 2), thendis even (otherwisea, b, c are all even) and s−p = db ≡ 0 (mod 4). In this case we haveps−qr ≡ 1−0 ≡ 1 (mod 4).
This contradictsps−qr = −1 which is equivalent to the condition N(εD) = −1.
Therefore, we have the possibilities p q
r s
≡ 1 0
1 1
, 1 1
0 1
, 0 1
1 0
(mod 2).
By Lemma 2.2, we seeS(α)6≡n (mod 2)and thusS(α)is even becausenis odd.
3. Proofs of Theorems
Proof of Theorem 1.1. Corresponding to the decomposition 8p= 2(xp+yp)2
+ 2(yp−xp)2
,
there exist exactly two elementsα, β ∈ Q(8p)which satisfyα=−1/α0,β =−1/β0. These are the largest roots of the equations
(3.1) (xp+yp)α2+ 2(xp−yp)α−(xp+yp) = 0
and (yp−xp)β2−2(xp+yp)β+xp−yp = 0 respectively. By our assumption on the norm of the fundamental unit (N(ε8p) =−1), we may takeαandβ as part of the representatives ofR(8p), the other representatives being in the formγ1,−1/γ10, . . . , γt,−1/γt0. From Lemma 2.3 (ii) we have
S(γi) +S(−1/γi) = 2S(γi)≡0 (mod 4).
Hence, by Proposition 2.1, it suffices to prove that S(α) +S(β)≡
( 0 (mod 4) if xpis even, 2 (mod 4) if xpis odd.
Let
α= [a0, . . . , an−1, an, an−1, . . . , a0] and β = [b0, . . . , bm−1, bm, bm−1, . . . , b0] be their (palindromic) continued fraction expansions (by our assumptionN(ε8p) =−1, their period lengths are odd) and set
p q r s
:=
a0 1 1 0
· · ·
an−1 1
1 0
,
P Q
R S
:=
p q r s
an 1 1 0
p r q s
, p0 q0
r0 s0
:=
b0 1 1 0
· · ·
bm−1 1
1 0
,
P0 Q0 R0 S0
:=
p0 q0 r0 s0
bm 1 1 0
p0 r0 q0 s0
.
Then we have
(3.2) Rα2+ (S−P)α−Q= 0, R0β2+ (S0−P0)β−Q0= 0. Setd:= GCD(R, S−P, Q)andd0= GCD(R0, S0−P0, Q0).
Lemma 3.1.
(i) P+S=P0+S0. (ii) d=d0.
Proof. LetεD be the fundamental unit of discriminantD. The classical construc- tion ofεD from the continued fraction expansions shows that the equalities
εD =Rα+S=R0β+S0 holds. ComputingεD +ε0Din two ways and using it as
εD+ε0D =R(α+α0) + 2S =R·P −S
R + 2S=P+S
=R0(β+β0) + 2S0=R0·P0−S0
R0 + 2S0 =P0+S0,
we obtain equality (i). The second equality (ii) follows similarly from the computation of(εD−ε0D)/√
D.
From (3.1) and (3.2), we have
R=Q=d(xp+yp), (3.3)
R0 =Q0 =d(yp−xp), (3.4)
P −S = 2d(yp−xp), (3.5)
P0−S0 = 2d(xp+yp).
(3.6)
By (3.3) and (3.4), we obtain
(3.7) R−R0=Q−Q0= 2dxp,
and by (3.5), (3.6) and Lemma 3.1 (i), we have
P−P0=S0−S=−2dxp.
By Lemma 2.3 (ii) and its proof, we havean≡bm≡0 (mod 2)and P Q
R S
≡
P0 Q0 R0 S0
≡ 0 1
1 0
(mod 2).
Then from (3.3), we have thatdis odd and from (3.5), we deduce that P−S≡2 (mod 4).
AssumeP ≡ 0, S ≡2 (mod 4). ByS =r2an+ 2rs≡ 2 (mod 4),r must be odd and we havean+ 2s≡ 2 (mod 4). Ifan ≡2 (mod 4), thensis even andq is odd. In this case, we have
Q=pran+qr+ps=pran+ 2qr+ (−1)n≡2p+ 2 + (−1)n (mod 4).
WhenQ ≡ 1 (mod 4),pis odd or even according asnis even or odd, and we have respectively
p q r s
≡ 1 1
1 0
(mod 2) or
p q r s
≡ 0 1
1 0
(mod 2).
In either case, we conclude by Lemma 2.2 thatS(α)≡2 (mod 4). Similarly we have S(α) ≡0 (mod 4)whenQ≡ −1 (mod 4). Also by the same argument we see that whenan≡0 (mod 4)we haveS(α)≡0 (mod 4)orS(α) ≡2 (mod 4)according asQ≡ −1 (mod 4)orQ≡1 (mod 4).
In summary, we have
S(α)≡
0 (mod 4) if
P Q
R S
≡
0 −1
−1 2
(mod 4),
2 (mod 4) if
P Q
R S
≡ 0 1
1 2
(mod 4).
Likewise, we obtain by considering the different cases and using Lemma 2.2 that
S(α)≡
0 (mod 4) if
P Q
R S
≡
2 −1
−1 0
(mod 4),
2 (mod 4) if
P Q
R S
≡ 2 1
1 0
(mod 4).
All these show that we have S(α)≡
( 0 (mod 4) if Q≡ −1 (mod 4), 2 (mod 4) if Q≡1 (mod 4).
Exactly the same holds forS(β), that is to say, S(β)≡
( 0 (mod 4) if Q0 ≡ −1 (mod 4), 2 (mod 4) if Q0 ≡1 (mod 4).
Now, by (3.7),Q≡Q0 (mod 4)if and only ifxp is even. Hence we have S(α) +S(β)≡
( 0 (mod 4) if xpis even, 2 (mod 4) if xpis odd,
which is what we need to establish Theorem 1.1.
Proof of Theorem 1.2. We proceed similarly, and keep the same notation. Letα andβbe the largest roots of the equations
xpxq+ypyq
2 α2− |xpyq−ypxq|α−xpxq+ypyq
2 = 0
and
ypyq−xpxq
2 β2−(xpyq+ypxq)β+xpxq−ypyq
2 = 0
respectively. These are the only elements inQ(pq)satisfyingα=−1/α0,β =−1/β0, corresponding to the decompositions
pq = (xpxq+ypyq)2+ (xpyq−ypxq)2
= (xpyq+ypxq)2+ (xpxq−ypyq)2.
By the assumptionxp 6≡ xq (mod 2), xpyq −ypxq and xpyq +ypxq are odd. As in the proof of Theorem 1.1, we may take as representatives ofR(pq) α, β, and the other in the form γ1,−1/γ10, . . . , γt,−1/γt0. By Lemma 2.3 (i) and the assumption N(εpq) =−1, we have
S(γi) +S(−1/γi) = 2S(γi)≡2 (mod 4).
By the theorem of Rédei-Reichardt [2],
]R(pq)≡0 (mod 4)(⇔tis odd)⇐⇒
q p
= 1.
Thus,
t
X
i=1
S(γi) +S(−1/γi0)
≡2 (mod 4)⇐⇒
q p
= 1, that is,
t
X
i=1
S(γi) +S(−1/γi0)
≡1 + q
p
(mod 4).
Since
κ+(pq) =S(α) +S(β) +
t
X
i=1
S(γi) +S(−1/γi0) by Proposition 2.1, to prove the theorem we need to show
S(α) +S(β) ≡ q
p
−1 + sgn(xpyq−ypxq)(−1)xp+l(α)−l(β)2
≡ 1−sgn(xpyq−ypxq)(−1)xp+l(α)−l(β)2 (mod 4).
(3.8)
(We have used the simple fact thatk≡ ±k (mod 4)whenkis even.)
Let the continued fraction expansions of α andβ be the same as in the proof of Theorem 1.1, and similarly construct the matrices
p q r s
,
P Q R S
, etc.
Then we have
R=Q=dxpxq+ypyq
2 ,
(3.9)
R0 =Q0 =dypyq−xpxq
2 ,
(3.10)
P−S =d|xpyq−ypxq|, (3.11)
P0−S0 =d(xpyq+ypxq), (3.12)
and (because Lemma 3.1 also holds in this case) R−R0=Q−Q0 =dxpxq, (3.13)
P−P0 =S0−S=−dmin(xpyq, ypxq). (3.14)
SinceD=pqis odd, by Lemma 2.3 (i) and Lemma 2.2 we have P Q
R S
≡ 1 0
0 1
or
1 1 1 0
or
0 1 1 1
(mod 2) andan≡bm ≡1 (mod 2).
We need to carry out rather tedious case distinctions, but here we describe how to deduce the conclusion only in the case when
P Q
R S
≡ 1 0
0 1
(mod 2), the other cases being similarly treated. In this case, because
Q=pran+ 2qr+ (−1)n≡0 (mod 2), pandrare both odd, and thus
P =p2an+ 2pq≡an+ 2q (mod 4).
SupposeP ≡ 1 (mod 4). Ifan ≡ 1 (mod 4), thenq is even and by Lemma 2.2 we concludeS(α) ≡ 1 (mod 4) (resp. ≡ 3 (mod 4)) whenn is odd (resp. even). If an ≡3 (mod 4), thenqis odd and we similarly haveS(α) ≡1 (mod 4)(resp. ≡3 (mod 4)) when nis odd (resp. even). Likewise, when P ≡ −1 (mod 4), we have S(α) ≡ 1 (mod 4)(resp. ≡ 3 (mod 4)) whennis even (resp. odd). Summing up, we conclude
S(α)≡
( (−1)n−1 (mod 4) if P ≡1 (mod 4), (−1)n (mod 4) if P ≡ −1 (mod 4).
Because QR ≡ 0 (mod 4) andP S −QR = −1, we have P 6≡ S (mod 4), whereupondnecessarily satisfiesd≡2 (mod 4)(notexpyq−ypxqis odd). By (3.13) and the assumptionxp6≡xq (mod 2), we haveR≡R0, Q≡Q0 (mod 4). Hence we
have
P0 Q0 R0 S0
≡ 1 0
0 1
(mod 2) and
S(β)≡
( (−1)m−1 (mod 4) if P0 ≡1 (mod 4), (−1)m (mod 4) if P0 ≡ −1 (mod 4).
Supposexpyq > ypxq,i.e.,sgn(xpyq−ypxq) = 1. Then by (3.14) P −P0=−dypxq ≡2ypxq≡1 + (−1)xp (mod 4) (recall the assumptionxp 6≡xq (mod 2)). From the above, we have,
S(α) +S(β)≡
( (−1)n+ (−1)m ≡ 1 + (−1)n−m (mod 4) ifxp is odd, (−1)n−(−1)m ≡ 1−(−1)n−m (mod 4) ifxp is even.
We can combine these into the congruence
S(α) +S(β)≡1−(−1)xp+n−m (mod 4).
Whenxpyq< ypxq, we have
P −P0=−dxpyq ≡2xpyq≡1−(−1)xp (mod 4) and in the same way we obtain
S(α) +S(β)≡1 + (−1)xp+n−m (mod 4).
We therefore have
S(α) +S(β)≡1−sgn(xpyq−ypxq)(−1)xp+n−m (mod 4).
Sincen−m= (l(α)−l(β))/2, we have (3.8) and Theorem 1.2 is proved.
Acknowledgement. The present work was partially supported by Grant-in-Aid for Scientific Research (B) No. 19340009.
R
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M. KANEKO, FACULTY OFMATH., KYUSHUU., 744 MOTOOKA, NISHI-KU, FUKUOKA819-0395, JAPAN
[email protected]-u.ac.jp
K. MORI, URESHINOSPECIALNEEDSEDUCATIONSCHOOL, GOCHOUDAKOU2877-1, SHIOTA, URE-
SHINO, SAGA849-1425, JAPAN
